Question

Ajudem nessa equação exponencial aqui ( 2/3) elevado a 2x+1 + (3/2) elevado a 2x-1 = 97 /54

Answer 1

$\left(\dfrac{2}{3} \right )^{\!\!2x+1}+\left(\dfrac{3}{2} \right )^{\!\!2x-1}=\dfrac{97}{54}\\\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x+1}+\left[\left(\dfrac{2}{3} \right )^{\!\!-1} \right ]^{\!2x-1}=\dfrac{97}{54}\\\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x+1}+\left(\dfrac{2}{3} \right )^{\!\!-1\,\cdot\, (2x-1)}=\dfrac{97}{54}\\\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x+1}+\left(\dfrac{2}{3} \right )^{\!\!-2x+1}=\dfrac{97}{54}\\\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x}\cdot \dfrac{2}{3}+\left(\dfrac{2}{3} \right )^{\!\!-2x}\cdot \dfrac{2}{3}=\dfrac{97}{54}$

$\dfrac{2}{3}\cdot \left[\left(\dfrac{2}{3} \right )^{\!\!2x}+\left(\dfrac{2}{3} \right )^{\!\!-2x} \right ]=\dfrac{97}{54}\\\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x}+\left(\dfrac{2}{3} \right )^{\!\!-2x}=\dfrac{97}{54}\cdot \dfrac{3}{2}\\\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x}+\left(\dfrac{2}{3} \right )^{\!\!-2x}=\dfrac{97}{\diagup\!\!\!\! 3\cdot 18}\cdot \dfrac{\diagup\!\!\!\! 3}{2}\\\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x}+\left(\dfrac{2}{3} \right )^{\!\!-2x}=\dfrac{97}{36}$

$\left(\dfrac{2}{3} \right )^{\!\!2x}+\dfrac{1}{\left(\frac{2}{3} \right )^{\!2x}}=\dfrac{97}{36}~~~~~~\mathbf{(i)}$


Faça a seguinte mudança de variável:

$\left(\dfrac{2}{3} \right )^{\!\!2x}=t~~~~(t>0)$


Substituindo, a equação $\mathbf{(i)}$ fica

$t+\dfrac{1}{t}=\dfrac{97}{36}\\\\\\ \dfrac{t^2}{t}+\dfrac{1}{t}=\dfrac{97}{36}\\\\\\ \dfrac{t^2+1}{t}=\dfrac{97}{36}\\\\\\ 36\cdot (t^2+1)=97t\\\\ 36t^2+36=97t\\\\ 36t^2-97t+36=0~~~~\Rightarrow~~\left\{ \!\begin{array}{l}a=36\\b=-97\\c=36 \end{array} \right.$

$\Delta=b^2-4ac\\\\ \Delta=(-97)^2-4\cdot 36\cdot 36\\\\ \Delta=9\,409-4\cdot 36\cdot 36\\\\ \Delta=9\,409-5\,184\\\\ \Delta=4\,225=65^2\\\\\\ t=\dfrac{-b\pm \sqrt{\Delta}}{2a}$

$t=\dfrac{-(-97)\pm \sqrt{65^2}}{2\cdot 36}\\\\\\ t=\dfrac{97\pm 65}{72}\\\\\\ \begin{array}{rcl} t=\dfrac{97-65}{72}&~\text{ ou }~&t=\dfrac{97+65}{72}\\\\\\ t=\dfrac{32}{72}\begin{array}{l}\,^{\div 8}\\\,^{\div 8} \end{array}&~\text{ ou }~&t=\dfrac{162}{72}\begin{array}{l}\,^{\div 18}\\\,^{\div 18} \end{array}\\\\\\ t=\dfrac{4}{9}&~\text{ ou }~&t=\dfrac{9}{4} \end{array}$


Substituindo de volta para a variável $x,$

$\begin{array}{rcl} \left(\dfrac{2}{3} \right )^{\!\!2x}=\dfrac{4}{9}&~\text{ ou }~&\left(\dfrac{2}{3} \right )^{\!\!2x}=\dfrac{9}{4}\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x}=\dfrac{2^2}{3^2}&~\text{ ou }~&\left(\dfrac{2}{3} \right )^{\!\!2x}=\dfrac{3^2}{2^2}\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x}=\left(\dfrac{2}{3} \right )^{\!\!2}&~\text{ ou }~&\left(\dfrac{2}{3} \right )^{\!\!2x}=\left(\dfrac{3}{2} \right )^{\!\!2}\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x}=\left(\dfrac{2}{3} \right )^{\!\!2}&~\text{ ou }~&\left(\dfrac{2}{3} \right )^{\!\!2x}=\left[\left(\dfrac{2}{3}\right)^{\!\!-1} \right ]^{2}\\\\ \left(\dfrac{2}{3} \right )^{\!\!2x}=\left(\dfrac{2}{3} \right )^{\!\!2}&~\text{ ou }~&\left(\dfrac{2}{3} \right )^{\!\!2x}=\left(\dfrac{2}{3}\right)^{\!\!-2}\\\\ \end{array}$


Agora, temos igualdades entre exponenciais de mesma base. Então, basta igualarmos os expoentes:

$\\\begin{array}{rcl} 2x=2&~\text{ ou }~&2x=-2\\\\ x=\dfrac{2}{2}&~\text{ ou }~&x=-\,\dfrac{2}{2} \end{array}\\\\\\ \quad\boxed{\begin{array}{rcl} x=1&~\text{ ou }~&x=-1 \end{array}}$


Conjunto solução: S = {– 1,  1}.


Bons estudos! :-)

Answer 2

$Ola'~~~ \\ \\ (\frac{2}{3}) ^{2x+1} +( \frac{3}{2}) ^{2x-1}= \frac{97}{54} \\ \\ \underline{Resolvendo~} \\ Usemos~algumas~propriedades~algebricas, temos: \\ \\ ( \frac{2}{3}) ^{2x}. \frac{2}{3}+( \frac{3}{2}) ^{2x}.( \frac{3}{2}) ^{-1}= \frac{97}{54}~~--\ \textgreater \ em~evidencia[ \frac{2}{3}] \\ \\ \frac{2}{3}\bigg[ ( \frac{2}{3} )^{2x}+( \frac{3}{2}) ^{2x}\bigg] = \frac{97}{54}~~--\ \textgreater \ simplificando~e~multiplicando \\ \\( \frac{2}{3} )^{2x}+ (\frac{3}{2}) ^{2x}= \frac{97}{36}$

$\frac{ 4^{x} }{ 9^{x} } + \frac{ 9^{x} }{ 4^{x} }= \frac{97}{36} \\ \\ \frac{ 4^{x}. 4^{x}+ 9^{x} . 9^{x} }{ 36^{x} } = \frac{4\¹.4\¹+9\¹.9\¹ }{36\¹ } \\ \\ Ent\~ao~: \\ 36^{x}=36\¹ \\ \\ \boxed{\boxed{x=1}} \\ \\ \\ \mathbb{iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~ter~ajudado!!$